Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace  std;
const int N=3e3+10;
double f[N][N],p[N];
int n;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%lf",&p[i]);
	f[0][0]=1,f[1][0]=1.0-p[1];
	double ans=0.000000000;
	for(int i=1;i<=n;i++)
	{
		for(int j=0;j<=i;j++)
		{
			if(!j)f[i][j]=f[i-1][j]*(1-p[i]);
			f[i][j]=f[i-1][j-1]*p[i]+f[i-1][j]*(1.0-p[i]);

		}
	}
	for(int i=n/2+1;i<=n;i++)ans+=f[n][i];
	printf("%.10f",ans);
    return 0;
}